Transmission Line Parameters ciflora 2022 Resistance • Power loss πΌ2 π • Voltage Drop affecting the voltage regulation of the line Resistance DC Resistance π ππ π =ρ π΄ where: π ππ = dc resistance, ohms ρ = resistivity of material at 20β, μΩ-cm π= length of conductor π΄=cross sectional area Resistance and temperature π 2 = π 1 1 + α π2 − π1 π 2 = temperature at π2 π 1 = temperature at π1 α = π‘πππππππ‘π’ππ πππππππππππ‘ ππ‘ 20β, β−1 Resistivity and Temperature Coefficient Resistivity (μΩ-cm) Temperature Coefficient (β−1 ) π΄ππ’ππππ’π = 2.83 πΆπππππ(βπππ ππππ€π) = 1.77 πΆπππππ(ππππππππ) = 1.72 π΄ππ’ππππ’π = 0.0039 πΆπππππ(βπππ ππππ€π) = 0.00382 πΆπππππ(ππππππππ) = 0.00393 Materials Used for Transmission Line Conductor Copper • high conductivity • high tensile strength • good ductility • Limitation is its cost Aluminium • sufficient conductivity • light in weight which results in low conductor weight and less sag • limitation is its low tensile strength • to overcome this limitation steel core is used to increase the tensile strength Types of overhead conductors used for overhead transmission and distribution • AAC : All Aluminum Conductor • AAAC : All Aluminum Alloy Conductor • ACSR : Aluminum Conductor, Steel Reinforced • ACAR : Aluminum Conductor, Alloy Reinforced Example Determine the resistance of a 10-km long solid cylindrical aluminum conductor with a diameter of a 250 mils at a. 20 β b. 120 β Example The per phase line loss in 40-km long transmission line is not to exceed 60 kW while it is delivering 100 A per phase. If the resistivity of the conductor material is 1.72 π₯ 10−8 Ωπ, determine the required conductor diameter. Exercise A single phase transmission line, 50-km long, is made up of hard drawn copper conductor 500 mils in diameter, find the resistance at a. 20 β. b. 80 β Inductance where: Single phase 2-wire line π·π = rπ One conductor πΏ = 2π₯10 −7 π· ln π· π πΏ = 4π₯10 π· ln π· π r=radius π· = πππ π‘ππππ πππ‘π€πππ πππππ’ππ‘πππ , H/m π· Two conductor −7 1 4 − , H/m Sample problem Find the inductive reactance per mile of a single phase line operating at 60 Hz. The conductor is Partridge and spacing is 20 ft between centers. Sample problem (alternate solution) Find the inductive reactance per mile of a single phase line operating at 60 Hz. The conductor is Partridge and spacing is 20 ft between centers. ππΏ = ππππ’ππ‘ππ£π πππππ‘ππ£π πππππ‘ππππ ππ 1 ππ‘ π ππππππ + ππππ’ππ‘ππ£π πππππ‘ππππ π ππππππ ππππ‘ππ ππΏ = ππ + ππ Inductance of composite conductors πΏ = 2π₯10 −7 π·π ln π· π , H/m where: π·π = πΊπππππ‘πππ ππππ π·ππ π‘ππππ (πΊππ·) π·π = πΊπππππ‘πππ ππππ π ππππ’π (πΊππ ) πΏ = πΏπ₯ + πΏπ¦ H/m π·π = πΊπππππ‘πππ ππππ π·ππ π‘ππππ (πΊππ·) π·π = πΊπππππ‘πππ ππππ π ππππ’π (πΊππ ) ππ.ππ ππππ‘πππ π·π = ππ.ππ ππππ‘πππ π·π = πππππ’ππ‘π ππ πππ πππ π‘πππππ πππ‘. π π‘πππππ ππ πππππ’ππ‘ππ π₯ π‘π πππ π π‘πππππ ππ πππππ’ππ‘ππ π¦ πππππ’ππ‘π ππ πππ πππ π‘πππππ πππ‘π€πππ π π‘πππππ ππ πππππ’ππ‘ππ πππππ’ππππ ππ‘π π π‘πππππ Example Example One circuit of a single phase TL is composed of three solid 0.25-cm radius wire. The return circuit id composed of two 0.5-cm radius wires. The arrangement is as shown. Find the inductance due to the current of the complete line. Inductance where: Three phase line π·ππ = πΏ = 2π₯10 ln π·ππ π·ππ π·ππ (depends on pole configuration) One conductor −7 3 π·ππ π·π , H/m/phase π·π = GMR Pole Configuration • Horizontal • Vertical • Triangular Example A single circuit 3-phase line operated at 60 Hz is arranged as shown. The conductors are Drake. Find the inductive reactance/mile 3 π·ππ = 20(20)(38) π·ππ = 24.8 ππ‘ from Table π·π = 0.0373 ππ‘ πΏ= πΏ= 24.8 2π₯10 ππ 0.0373 π» 13π₯10−3 π −7 ππΏ = 2π 60 13π₯10 −3 Ω ππΏ = 0.788 /πβππ π ππ 1609 π 1 ππ Alternate solution: Using tables ππΏ = ππ + ππ ππ = 0.399 ππ = 0.389 ππΏ = 0.3999 + 0.389 Ω ππΏ = 0.788 /πβππ π ππ Example A 30-km 3-phase transmission line operated at 60 Hz is arranged as shown. Find the total inductive reactance/mile of the whole line. What will be the total inductive reactance if the pole configuration is horizontal? Bundled Conductors • combination of more than one conductor per phase in parallel suitably spaced from each other used in an overhead transmission line • used to increase electrical capacity and reduce corona and radio noise at voltages above 200 kV. Bundled Conductors… For a 2-strand bundle π π·π = 4 π·π π For a 4-strand bundle π·π π = 1.09 2 For a 3-strand bundle π π·π = 9 π·π π π π·π π 3 where: 3 π = πππ π‘ππππ πππ‘π€πππ π π‘ππππ Example Each conductor of the bundled conductor shown in figure is ACSR, 1,272,000 cmil Pheasant. Find the inductive reactance in ohms/km (and per mile) per phase for d=45 cm. Find the per unit series reactance of the line if its length is 160 km and base is 100 MVA at 345 kV. Ds for Pheasant is 0.0466 ft. Double circuit three phase line Single circuit three phase line Double circuit three phase line Double circuit three phase line • Two circuit • Enables the transfer of more power over a particular distance π π·ππ = π π·π = 3 3 π·ππ π π·ππ π π·ππ π π·ππ′ π π·ππ′ π π·ππ′ π Example A 3Π€ double circuit line is composed of 300,000 cmil 26/7 ACSR Ostrich conductors as shown. Find the 60 Hz inductive reactance in ohms/mi per phase. Ds for Ostrich conductors is 0.0229 ft. Transmission Line Parameters 2 ciflora 2022 Capacitance where: Single phase 2-wire line πΆππ = πππππππ‘ππππ πππ‘π€πππ ππππ π πππ π π· = πππ π‘ππππ πππ‘π€πππ πππππ’ππ‘πππ ππ = ππ’π‘π πππ πππππ’π ππ πππππ’ππ‘ππ π πΆππ = 2ππ0 π·2 ππ ππ ππ ππ = ππ’π‘π πππ πππππ’π ππ πππππ’ππ‘ππ π F/m π0 = ππππππ‘ππ£ππ‘π¦ ππ ππππ π ππππ = 1π₯10−9 36π Capacitance If ππ = ππ πΆππ = Three phase line 2ππ0 π· 2ππ π F/m Line-to-neutral capacitance πΆπ = 2ππ0 π· ππ π F/m to neutral πΆπ = 2ππ0 π·ππ ππ π F/m to neutral Use of Table ππ = πππππππ‘ππ£π πππππ‘ππππ ππ 1 ππ‘ π ππππππ + πππππππ‘ππ£π πππππ‘ππππ π ππππππ ππππ‘ππ ππ = π′π + ππ Bundled Conductors For a 2-strand bundle π π·π = 4 ππ For a 4-strand bundle π 2 4 π·π = 1.09 ππ3 where: π = ππ’π‘π πππ πππππ’π ππ πππππ’ππ‘ππ For a 3-strand bundle π π·π = 9 πππ 3 Double circuit lines • replace GMR or Ds by outside radius of conductor Example Find the capacitive susceptance per mile of a single phase line operating at 60 Hz. The conductor is Partridge and spacing is 20 ft between centers. Example Find the capacitive reactance for 1 mile of a single circuit 3Π€ line operated at 60 Hz and arranged as shown. The conductors are ACSR Drake. If the length of the line is 175 miles, find the capacitive reactance to neutral for the entire length of the line. Example Find the capacitive reactance to neutral of the line shown. Each conductor of the bundled conductor line shown is ACSR Pheasant. TABLE A.3 Electrical characteristics of bare aluminum conductors steel-reinforced (ACSR)* Resistance Ac, 60 Hz Reactance per conductor 1-ft spacing, 60 Hz Code word 20°C, Ω/mi 50°C, Ω/mi GMR Ds ft Inductive Xa, Capacitive X'a, Ω/mi MΩ.mi Aluminum area, Stranding Layers of Outside diameter, Dc, 20°C, cmil Al/St aluminum in Ω/1,000ft Waxwing 266.800 18/1 2 0,609 0,0646 0,3488 0,3831 0,0198 0,476 0,1090 Partridge 266.800 26/7 2 0,642 0,0640 0,3452 0,3792 0,0217 0,465 0,1074 Ostrich 300.000 26/7 2 0,680 0,0569 0,3070 0,3372 0,0229 0,458 0,1057 Merlin 336.400 18/1 2 0,684 0,0512 0,2767 0,3037 0,0222 0,462 0,1055 Linnet 336.400 26/7 2 0,721 0,0507 0,2737 0,3006 0,0243 0,451 0,1040 Oriole 336.400 30/7 2 0,741 0,0504 0,2719 0,2987 0,0255 0,445 0,1032 Chickadee 397.500 18/1 2 0,743 0,0433 0,2342 0,2572 0,0241 0,452 0,1031 Ibis 397.500 26/7 2 0,783 0,0430 0,2323 0,2551 0,0264 0,441 0,1015 Pelican 477.000 18/1 2 0,814 0,0361 0,1957 0,2148 0,0264 0,441 0,1004 Flicker 477.000 24/7 2 0,846 0,0359 0,1943 0,2134 0,0284 0,432 0,0992 Hawk 477.000 26/7 2 0,858 0,0357 0,1931 0,2120 0,0289 0,430 0,0988 Hen 477.000 30/7 2 0,883 0,0355 0,1919 0,2107 0,0304 0,424 0,0980 Osprey 556.500 18/1 2 0,879 0,0309 0,1679 0,1843 0,0284 0,432 0,0981 Parakeet 556.500 24/7 2 0,914 0,0308 0,1669 0,1832 0,0306 0,423 0,0969 Dove 556.500 26/7 2 0,927 0,0307 0,1663 0,1826 0,0314 0,420 0,0965 Rook 636.000 24/7 2 0,977 0,0269 0,1461 0,1603 0,0327 0,415 0,0950 Grosbeak 636.000 26/7 2 0,990 0,0268 0,1454 0,1596 0,0335 0,412 0,0946 Drake 795.000 26/7 2 1,108 0,0215 0,1172 0,1284 0,0373 0,399 0,0912 Tern 795.000 45/7 3 1,063 0,0217 0,1188 0,1302 0,0352 0,406 0,0925 Rail 954.000 45/7 3 1,165 0,0181 0,0997 0,1092 0,0386 0,395 0,0897 Cardinal 954.000 54/7 3 1,196 0,0180 0,0988 0,1082 0,0402 0,390 0,0800 Ortolan 1.033.500 45/7 3 1,213 0,0167 0,0924 0,1011 0,0402 0,390 0,0885 Bluejay 1.113.000 45/7 3 1,259 0,0155 0,0861 0,0941 0,0415 0,386 0,0874 Finch 1.113.000 54/19 3 1,293 0,0155 0,0856 0,0937 0,0436 0,380 0,0866 Bittern 1.272.000 45/7 3 1,345 0,0136 0,0762 0,0832 0,0444 0,378 0,0855 Pheasant 1.272.000 54/19 3 1,382 0,0135 0,0751 0,0821 0,0466 0,372 0,0847 Bobolink 1.431.000 45/7 3 1,427 0,0121 0,0684 0,0746 0,0470 0,371 0,0837 Plover 1.431.000 54/19 3 1,465 0,0120 0,0673 0,0735 0,0494 0,365 0,0829 Lapwing 1.590.000 45/7 3 1,502 0,0109 0,0623 0,0678 0,0498 0,364 0,0822 Falcon 1.590.000 54/19 3 1,545 0,0108 0,0612 0,0667 0,0523 0,358 0,0814 Bluebird 2.156.000 84/19 4 1,762 0,0080 0,0476 0,0515 0,0586 0,344 0,0776 * Most used multilayer sizes. ** Data, by permission, from Aluminum Association, Aluminum Electrical Conductor Handbook, 2nd ed., Washington, D.C., 1982. Copiado de: J.J.Grainger, W.D.Stevenson, Power System Analysis, McGraw-Hill, 1994. -1/5- TABLE A.4 Inductive reactance spacing factor Xd at 60 Hz* (ohms per mile per conductor) Separation Inches Feet 0 1 2 3 4 5 6 7 8 9 10 11 0 ...... -0,3015 -0,2174 -0,1682 -0,1333 -0,1062 -0,0841 -0,0654 -0,0492 -0,0349 -0,0221 -0,0106 1 0,0000 0,0097 0,0187 0,0271 0,0349 0,0423 0,0492 0,0558 0,0620 0,0679 0,0735 0,0789 2 0,0841 0,0891 0,0938 0,0984 0,1028 0,1071 0,1112 0,1152 0,1190 0,1227 0,1264 0,1299 3 0,1333 0,1366 0,1399 0,1430 0,1461 0,1491 0,1520 0,1549 0,1577 0,1604 0,1631 0,1657 4 0,1682 0,1707 0,1732 0,1756 0,1779 0,1802 0,1825 0,1847 0,1869 0,1891 0,1912 0,1933 5 0,1953 0,1973 0,1993 0,2012 0,2031 0,2050 0,2069 0,2087 0,2105 0,2123 0,2140 0,2157 6 0,2174 0,2191 0,2207 0,2224 0,2240 0,2256 0,2271 0,2287 0,2302 0,2317 0,2332 0,2347 7 0,2361 0,2376 0,2390 0,2404 0,2418 0,2431 0,2445 0,2458 0,2472 0,2485 0,2498 0,2511 8 0,2523 9 0,2666 10 0,2794 11 0,2910 12 0,3015 13 0,3112 14 0,3202 I5 0,3286 16 0,3364 17 0,3438 18 0,3507 19 0,3573 20 0,3635 21 0,3694 22 0,3751 23 0,3805 24 0,3856 At 60 Hz, in Ω/mi per conductor 25 0,3906 Xd = 0.2794 log d 26 0,3953 d= separation, ft 27 0,3999 For three-phase lines 28 0,4043 d= Deq 29 0,4086 30 0,4127 31 0,4167 32 0,4205 33 0,4243 34 0,4279 35 0,4314 36 0,4348 Copiado de: J.J.Grainger, W.D.Stevenson, Power System Analysis, McGraw-Hill, 1994. -2/5- 37 0,4382 38 0,4414 39 0,4445 40 0,4476 41 0,4506 42 0,4535 43 0,4564 44 0,4592 45 0,4619 46 0,4646 47 0,4672 48 0,4697 49 0 4722 * From Electrical Transmission and Distribution Reference Book, by permission of the ABB Power T & D Company, Inc. Copiado de: J.J.Grainger, W.D.Stevenson, Power System Analysis, McGraw-Hill, 1994. -3/5- TABLE A.5 Shunt capacitance-reactance spacing factor Xd at 10 Hz (megaohm-miles per conductor) Separation Inches Feet 0 1 2 3 4 5 6 7 8 9 10 11 0 ...... -0,0737 -0,0532 -0,0411 -0,0326 -0,0260 -0,0206 -0,0160 -0,0120 -0,0085 -0,0054 -0,0026 1 0,0000 0,0024 0,0046 0,0066 0,0085 0,0103 0,0120 0,0136 0,0152 0,0166 0,0180 0,0193 2 0,0206 0,0218 0,0229 0,0241 0,0251 0,0262 0,0272 0,0282 0,0291 0,0300 0,0309 0,0318 3 0,0326 0,0334 0,0342 0,0350 0,0357 0,0365 0,0372 0,0379 0,0385 0,0392 0,0399 0,0405 4 0,0411 0,0417 0,0423 0,0429 0,0435 0,0441 0,0446 0,0452 0,0457 0,0462 0,0467 0,0473 5 0,0478 0,0482 0,0487 0,0492 0,0497 0,0501 0,0506 0,0510 0,0515 0,0519 0,0523 0,0527 6 0,0532 0,0536 0,0540 0,0544 0,0548 0,0552 0,0555 0,0559 0,0563 0,0567 0,0570 0,0574 7 0,0577 0,0581 0,0584 0,0588 0,0591 0,0594 0,0598 0,0601 0,0604 0,0608 0,0611 0,0614 8 0,0617 9 0,0652 10 0,0683 11 0,0711 12 0,0737 13 0,0761 14 0,0783 I5 0,0803 16 0,0823 17 0,0841 18 0,0858 19 0,0874 20 0,0889 21 0,0903 22 0,0917 23 0,0930 24 0,0943 At 60 Hz, in MΩ.mi per conductor 25 0,0955 Xd' = 0.06831 log d 26 0,0967 d= separation, ft 27 0,0978 For three-phase lines 28 0,0989 d= Deq 29 0,0999 30 0,1009 31 0,1019 32 0,1028 33 0,1037 34 0,1046 35 0,1055 36 0,1063 Copiado de: J.J.Grainger, W.D.Stevenson, Power System Analysis, McGraw-Hill, 1994. -4/5- 37 0,1071 38 0,1079 39 0,1087 40 0,1094 41 0,1102 42 0,1109 43 0,1116 44 0,1123 45 0,1129 46 0,1136 47 0,1142 48 0,1149 49 0,1155 * From Electrical Transmission and Distribution Reference Book, by permission of the ABB Power T & D Company, Inc. Copiado de: J.J.Grainger, W.D.Stevenson, Power System Analysis, McGraw-Hill, 1994. -5/5- Short Transmission Line effective length less than 80 km (50 miles) ciflora 2022 Parameters • Series resistance • Series inductance • Shunt capacitance (neglected) • Shunt conductance (neglected) Assumptions • System is Y-connected • Transmission is balance Equivalent Circuit πΌπ = πΌπ ππ = ππ + πΌπ π πππ = πΆππ ππΌπ − πππ πππΏ − ππΉπΏ %ππ = π₯ 100 ππΉπΏ πππΏ − ππΉπΏ %ππ· = π₯ 100 πππΏ Equivalent Circuit… where: πΌπ = π ππππππ πππ ππ’πππππ‘ πΌπ = ππππππ£πππ πππ ππ’πππππ‘ ππ = π ππππππ πππ ππππ π£πππ‘πππ πππ = π ππππππ πππ πβππ π π£πππ‘πππ ππ = ππππππ£πππ πππ ππππ π£πππ‘πππ ππ π = ππππππ£πππ πππ πβππ π π£πππ‘πππ πππ = π ππππππ πππ πππ€ππ ππππ‘ππ πππ = ππππππ£πππ πππ πππ€ππ ππππ‘ππ %ππ = πππππππ‘ π£πππ‘πππ ππππ’πππ‘πππ %ππ· = πππππππ‘ π£πππ‘πππ ππππ ππ = π ππππππ πππ πππ€ππ ππππ‘ππ πππππ ππ = ππππππ£πππ πππ πππ€ππ ππππ‘ππ πππππ Phasor diagram ππ = ππ + πΌπ π VR and VD • Voltage regulation is a measure of how much voltage is dropped along the length of the transmission line from the sending end to the receiving end Example An overhead, 1-phase transmission line delivers 1100 kW at 13.8 kV at 0.78 pf lagging. The resistance and reactance per wire is 5 ohms and 7.5 ohms respectively. Calculate: a. Sending end voltage and pf b. Transmission line efficiency Example… • A 3-phase 10 mi distribution feeder 3-336.4 MCM 26/7 strand ACSR are spaced 3 ft. horizontally on steel tower. It is supplying a load of 5000 kW at 0.8 pf lag 13.2 kV. For 336.4 MCM ACSR, GMR at 60 Hz is 0.0244 ft and resistance at 60 Hz. 50°C is 0.306 ohm/mi. Calculate a. Impedance of the line b. Sending end voltage and pf c. % voltage drop Example A 3-phase transmission line 25 km long is supplying power to a small town that draws 75 A at 86.6% pf lag 69 kV. The substation supplying the line has an output voltage of 72.25 kV at 85.1% pf lag. Calculate the resistance and reactance of line. Example… A 3-phase transmission line50 miles long has an impedance of 5+j40 ohms per phase. The terminal voltages at both ends are 345 kV and 360 kV. Assume that sending end voltage leads receiving end voltage by 10 degrees. Compute: a. Real and reactive powers in each end b. Total line loss, P and Q. 1. During the interview of DOE Undersecretary Gerardo Erguiza this morning, regarding the continuing oil price increase, he mentioned that the agency is looking into the details of unbundling of energy cost. One of the provisions of EPIRA is the unbundling of electricity retail tariffs. Explain this provision (maximum of 2 sentences) Electricity charge is sub categorized into several charges (generation, transmission, distribution, etc.) instead of 1 to 2 categories (electricity cost, price cost adjustment) only 2. Find the GMR of the unconventional conductors shown if the individual strand has a diameter of 2 cm. πΊππ = 9√π·ππ π·ππ π·ππ π·ππ π·ππ π·ππ π·ππ π·ππ π·ππ = π. πππ ππ 1 π·ππ = π·ππ = π·ππ = (1)π −4 = 0.7788 ππ π·ππ = π·ππ = π·ππ = π·ππ = 2 π·ππ = π·ππ = 4 3. Calculate the inductive reactance of a double-circuit, 3-phase, transposed 80-km transmission line as shown if the radius of each conductor is 1.25 cm. π» 1000 π π» ( ) = 0.607π₯10−3 π 1 ππ ππ π» ππΏ = 2π(60) (0.607π₯10−3 ) (80 ππ) = 18.31 πβππ ππ πΏ = 0.607π₯10−6 A 3-phase short transmission line with an impedance of 2 + j5 ohms supplies the following loads at 8000 V to neutral: an inductive load 3000 kVAR at 0.8 pf, and a capacitive load of 600 kVAR at 0 pf. Calculate a. sending end voltage and pf b. line loss ππΏ = πππ −1 0.8 = 36.87° ππΏ ππ = ππΏ + ππ = + 0 = 4000 ππ tan 36.87 ππ = ππΏ − ππΆ = 3000 − 600 = 2400 πππ΄π (πππ. ) ππ 2400 ππ = π‘ππ−1 = π‘ππ−1 = 31° ππ 4000 4000π₯103 πΌπ = = 194.4 ∠ − 31 π΄ 3(8000)(cos 31) or 3000π₯103 πΌπΏ = ∠ − 36.87 = 208.3∠ − 36.87 π΄ 3 (8000) sin 36.87 600π₯103 πΌπ = ∠90 = 25∠90 π΄ 3 (8000) sin 90 πΌπ = πΌπΏ + πΌπ = 194.4 ∠ − 31 π΄ πππ = 8000∠0 + 194.4∠ − 31(2 + π5) = 8856.5∠4 ππ = √3(8856.5) = ππ. ππ ππ½ πππ = πΆππ |−31 − 4| = π. πππ πππ ππππ π = 3πΌπ 2 π = 3(194.4)2 (2) = πππ. π ππΎ An 18 km 60 Hz 3-phase transmission line of conductor resistance of 0.3792 ohms/mi, inductive reactance of 1- ft spacing of 0.465 ohms/mi, and inductive reactance spacing factor of 0.2012 ohms/mi are equilaterally spaced with 1.6 m between centers. The line delivers 2500 kW at 11 kV to a balanced load. Determine the percent regulation if the load has 90% leading pf. πππ = ππ π + πΌπ π 2500 πΌπ = πΌπ = = 145.79 ∠ cos −1 0.9 = 145.79 ∠25.84 π΄ √3(11)(0.9) πβππ π = 0.3792 = 4.24 πβππ ππ πβππ 1 ππ ππΏ = (0.465 + 0.2012) ( ) (18 ππ) = 7.45 πβππ ππ 1.609 ππ 11π₯103 πππ = ∠0 + 145.79 ∠25.84(4.24 + π7.45) = 6553.52 ∠10.97 √3 ππ = √3(6553.52 ) = 11.35 ππ 11.35 − 11 %ππ = π₯100 = π. ππ% 11 A 3-phase 60 Hz line has flat horizontal spacing. The conductors have an outside diameter of 3.28 cm with 12 m between conductors. Determine the capacitive reactance of the line in ohms if its length is 125 mi. 3 π·ππ = √12(12)(24) = 15.12 π 0.0328 π= = 0.0164 π 2 1 πβπ π 1 ππ ππ = ( ) = ππππ. π ππππ 125 ππ 1609 π 2πππ 2π(60) ( ) 15.12 ππ 0.0164 Medium Transmission Line effective length less than 81-240 km (150 miles) ciflora 2022 Parameters • Series resistance • Series inductance • Shunt capacitance • Shunt conductance (neglected) Representation • Nominal T-circuit Representation • Nominal π-circuit Representation Nominal T-Circuit Representation ππ = ππ π + πΌπ ππΏ 2 πΌπ = ππ π πΌπ = πΌπ + πΌπ πππ = ππΆ + πΌπ %ππ = ππΏ 2 πππΏ − ππΉπΏ π₯ 100 ππΉπΏ πππΏ − ππΉπΏ %ππ· = π₯ 100 πππΏ πππΏ πππ ππ ππ = ππ 1+ 2 Example A 3-phase transmission line has the following data: length is 96 km, conductors are spaced 2.5 m horizontally. It is supplying a balance load drawing 200 A at 0.85% pf lagging, 230 kV line-to-line and 60 Hz. The line uses 500 MCM ACSR, GMR of 0.0311 ft and the resistance at 60 Hz 50°C is 0.206 ohm/mi and the outside radius is 0.452 in. Assuming that the capacitance are lumped at the middle of the line and that the wire temperature is 50°C. Calculate a. Sending end current b. Sending end voltage c. %VR d. Efficiency Nominal π-Circuit Representation ππΆπ = ππ π π πΌπΆπ = ππ π 2 πΌπ = πΌπ + πΌπΆπ ππΆπ = ππΆπ + πΌπ§ π πππ = ππΆπ π πΌπΆπ = ππΆπ 2 πΌπ = πΌπ§ + πΌπΆπ πππΏ − ππΉπΏ %ππ = π₯ 100 ππΉπΏ %ππ· = πππΏ − ππΉπΏ π₯ 100 πππΏ πππΏ = πππ ππ ππ ππ 1+ 2 Example A 3-phase transmission line has the following constant: impedance per wire is 15+j20 ohms, shunt susceptance at the middle of the line is 0.0025 mhos from line to neutral, receiving end voltage is 69 kV and the receiving kVA is 25000 at 80% pf lag. Calculate: a. %VR b. Efficiency Long Transmission Line effective length more than 240 km (150 miles) ciflora 2022 Parameters • Series resistance • Series inductance • Shunt capacitance • Shunt conductance Representation Voltage Equation πππ = ππ π cosh ππ + πΌπ ππΆ sinh ππ Current Equation ππ π πΌπ = πΌπ cosh ππ + sinh ππ ππΆ where ππΆ = πβπππππ‘ππππ π‘ππ ππππππππ ππΆ = π§ π Example A 3-phase 175 miles long transmission line is supplying a balance load of 150 MW at 80% pf lagging 230 kV. The line has the following constants: x=0.6 ohms/mi, r=0.204 ohms/mi, y=6 micro S/mi. Find sending end voltage. Transmission line as 2-port network • Represented by 4-terminal network with ABCD constants • The ABCD parameters of a transmission line give the relationship of the input voltage and currents to the output voltage and currents. • ABCD parameters simplify complex calculations when transmission lines are cascaded. • ABCD parameters are dependent on the length of a transmission line. ππ = π΄ππ + π΅πΌπ πΌπ = πΆππ + π·πΌπ ABCD Constants for Transmission Lines (per phase) Line Length Equiv. Circuit A B C D Short Series Impedance π π Medium Nominal π π ππ π+ π π ππ π+ π Nominal T ππ π+ π ππ π π+ π π ππ π+ π Distributed ππ¨π¬π‘ ππ ππͺ π¬π’π§π‘ ππ ππππ ππ ππͺ ππ¨π¬π‘ ππ Long π π π+ ππ π Example For a 3-phase long transmission line Zc= 406.4∠-5.48 Ω, Vr=215 kV and Ir=335.7 A at unity pf. Evaluate the ABCD constants and calculate the sending end voltage if Y=3.2 μS Electrical Fault ciflora 2022 Electrical Fault • deviation of voltages and currents from nominal values or states • under normal operating conditions, power system equipment or lines carry normal voltages and currents which results in safer operation of the system • when a fault occurs, it causes excessively high currents to flow which causes damage to equipment and devices Causes • Lightning • Heavy winds • Trees falling across lines • Vehicles colliding with towers or poles • Aircraft colliding with lines • Vandalism • Line breaks due to excessive loading Types Open Circuit Fault (also called series fault) Short Circuit Fault (also called shunt fault) • conductors of the different phases come into contact with each other with a power line, power transformer or any other circuit element due to which the large current flow in one or two phases of the system. • divided into the symmetrical and unsymmetrical fault. Symmetrical Fault • The faults which involve all the three phases • Uncommon, but most severe • Three phase fault • Three phase to ground fault Unsymmetrical Fault • gives rise to unsymmetrical current, i.e., current differing in magnitude and phases in the three phases • Single Line to Ground • Line to Line • Double Line to Ground Calculations π = 3ππΏ πΌπΏ Assumption Y-connection, πΌπ = πΌπΏ πΌπ = πΌπΏ = πππ΄ π₯ 103 3πΎππΏ π΄ππ πΎππ₯ 103 2 ππ πΎπ 3 ππ = = = πβππ 3 πΌπ πππ΄ π₯ 10 πππ΄ 3πΎππΏ Calculations… π΄ππ‘π’ππ ππππ’π πππ ππππ‘ ππππ’π = π΅ππ π ππππ’π ππππ π πΌπππ π = πππ’ πππ€ πΎπ 2 πππ π = πππ΄πππ π πππ΄πππ π π₯ 103 3πΎππππ π %ππππ£ππ πΎππππ£ππ = 100% πΎππππ π 2 πππ΄πππ π πππ΄πππ£ππ Example Calculate the fault current at point F if a 3 phase fault occurs at point F Exercise Calculate the fault current at point F if a 3 phase fault occurs at point F
